3.812 \(\int \frac{(a+b \tan (c+d x))^2}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=268 \[ \frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)} \]

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b^2)/(7*d*Cot[c + d*x]^(7/2)) + (4*a*b)/(5*d*Cot[c + d*x]^(5/2))
+ (2*(a^2 - b^2))/(3*d*Cot[c + d*x]^(3/2)) - (4*a*b)/(d*Sqrt[Cot[c + d*x]]) - ((a^2 + 2*a*b - b^2)*Log[1 - Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*
x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.285769, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {3673, 3542, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^2/Cot[c + d*x]^(5/2),x]

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b^2)/(7*d*Cot[c + d*x]^(7/2)) + (4*a*b)/(5*d*Cot[c + d*x]^(5/2))
+ (2*(a^2 - b^2))/(3*d*Cot[c + d*x]^(3/2)) - (4*a*b)/(d*Sqrt[Cot[c + d*x]]) - ((a^2 + 2*a*b - b^2)*Log[1 - Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*
x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3542

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^2}{\cot ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{(b+a \cot (c+d x))^2}{\cot ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\int \frac{2 a b+\left (a^2-b^2\right ) \cot (c+d x)}{\cot ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\int \frac{a^2-b^2-2 a b \cot (c+d x)}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\int \frac{-2 a b-\left (a^2-b^2\right ) \cot (c+d x)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\int \frac{-a^2+b^2+2 a b \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{a^2-b^2-2 a b x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}+\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}+\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 d}+\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 d}-\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{2 \sqrt{2} d}\\ &=\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}\\ &=-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{2 b^2}{7 d \cot ^{\frac{7}{2}}(c+d x)}+\frac{4 a b}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (a^2-b^2\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{4 a b}{d \sqrt{\cot (c+d x)}}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}\\ \end{align*}

Mathematica [C]  time = 0.392241, size = 80, normalized size = 0.3 \[ \frac{2 \left (a \left (5 a+14 b \cot (c+d x) \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\cot ^2(c+d x)\right )\right )-5 \left (a^2-b^2\right ) \, _2F_1\left (-\frac{7}{4},1;-\frac{3}{4};-\cot ^2(c+d x)\right )\right )}{35 d \cot ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^2/Cot[c + d*x]^(5/2),x]

[Out]

(2*(-5*(a^2 - b^2)*Hypergeometric2F1[-7/4, 1, -3/4, -Cot[c + d*x]^2] + a*(5*a + 14*b*Cot[c + d*x]*Hypergeometr
ic2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2])))/(35*d*Cot[c + d*x]^(7/2))

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Maple [C]  time = 0.362, size = 1903, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^2/cot(d*x+c)^(5/2),x)

[Out]

-1/210/d*2^(1/2)*(cos(d*x+c)-1)*(105*I*sin(d*x+c)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2
-1/2*I,1/2*2^(1/2))*b^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(
d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-210*I*cos(d*x+c)^3*sin(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b-105*I*sin(d*x+c)*EllipticPi((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1
+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+210*I*cos(d*x+c)^3*s
in(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d
*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b-1
05*I*sin(d*x+c)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2*((cos(d*x+
c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(
1/2)*cos(d*x+c)^3+105*I*sin(d*x+c)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1
/2))*a^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+105*cos(d*x+c)^3*sin(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin
(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a^2+210*cos(d*x+c)^3*sin(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b-105*cos(d*x+c)^3*sin(d*x+c)*((cos(d*x+c
)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1
/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2+105*cos(d*x+c)^3*sin(d
*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c
))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a^2+210*c
os(d*x+c)^3*sin(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2
^(1/2))*a*b-105*cos(d*x+c)^3*sin(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2
),1/2+1/2*I,1/2*2^(1/2))*b^2-420*sin(d*x+c)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2)*cos(d*x+c)^3*a*b+504*2^(1/2)*cos(d*x+c)^4*a*b-70*sin(d*x+c)*2^(1/2)*a^2*cos(d*x+c)^3+100*sin
(d*x+c)*2^(1/2)*b^2*cos(d*x+c)^3-504*2^(1/2)*cos(d*x+c)^3*a*b+70*sin(d*x+c)*2^(1/2)*a^2*cos(d*x+c)^2-100*sin(d
*x+c)*2^(1/2)*b^2*cos(d*x+c)^2-84*2^(1/2)*cos(d*x+c)^2*a*b-30*2^(1/2)*b^2*cos(d*x+c)*sin(d*x+c)+84*2^(1/2)*cos
(d*x+c)*a*b+30*2^(1/2)*b^2*sin(d*x+c))*(cos(d*x+c)+1)^2/(cos(d*x+c)/sin(d*x+c))^(5/2)/sin(d*x+c)^6/cos(d*x+c)

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Maxima [A]  time = 1.69309, size = 302, normalized size = 1.13 \begin{align*} \frac{8 \,{\left (15 \, b^{2} + \frac{42 \, a b}{\tan \left (d x + c\right )} - \frac{210 \, a b}{\tan \left (d x + c\right )^{3}} + \frac{35 \,{\left (a^{2} - b^{2}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{7}{2}} + 210 \, \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 210 \, \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 105 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - 105 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2/cot(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/420*(8*(15*b^2 + 42*a*b/tan(d*x + c) - 210*a*b/tan(d*x + c)^3 + 35*(a^2 - b^2)/tan(d*x + c)^2)*tan(d*x + c)^
(7/2) + 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 210*sqrt(2)*(a^
2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(s
qrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)/sqrt(tan(d*x +
c)) + 1/tan(d*x + c) + 1))/d

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2/cot(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**2/cot(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}{\cot \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^2/cot(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^2/cot(d*x + c)^(5/2), x)